Question: Rewrite the function by completing the square. $f(x)=x^{2}-14x+63$ $f(x)=(x+$
We want to complete $x^2{-14}x$ into a perfect square. To do that, we should add $\left(\dfrac{{-14}}{2}\right)^2={49}$ to it: $x^2{-14}x+{49}=(x-7)^2$ In order to keep the expression equivalent, we add and subtract ${49}$, not forgetting the expression's constant term, $63$ : $\begin{aligned} f(x)&=x^2-14x+63 \\\\ &=x^2-14x+{49}+63-{49} \\\\ &=(x-7)^2+63-49 \\\\ &=(x-7)^2+14 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x - 7)^2 + 14$ This is equivalent to $f(x)=(x+{-7})^2+14$